parameter undefined in procedure

[Moshe Zadka]

On Fri, 25 Feb 2000, David Smith wrote:

> I have a module named test.py, which contains only the following
> definition:
> 
> def bar(dictionary):
>   list = ['alpha', 'bravo', 'charlie', 'delta', 'echo', 'foxtrot']
>   return map( lambda(x): dictionary.get(x,"default"), list)
> 
> I import it and run it, and get an error message as follows:
<a name error on dictionary>

When a Python function sees a variable, it looks in three places (in that
order) for the variable: the local namespace, the global namespace, and
the builtin namespace.

The "lambda" expression defines a function. Inside it, the only locals
are the parameters (since there can be no assignments inside a lambda),
so in your case the only local is "x". "dictionary" isn't a global,
and it isn't a builtin (thank God!) either. If you want your lambda
expression to see it, you have to explicitly define it:

return map(lambda x, dictionary=dictionary: dictionary.get(x, "default"),
            list)

If you think this is god-awful ugly, you're right. Why not code it 
as an explicit loop?
--
Moshe Zadka <mzadka at geocities.com>. 
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